Issue 29

S. Terravecchia et al., Frattura ed Integrità Strutturale, 29 (2014) 61-73; DOI: 10.3221/IGF-ESIS.29.07 67 , 0 , 0 x x x y y y u u c1 g n u u c2 g n             (19a,b,c,d) with c1 and 2 c being the wideness of the displacements; - for the case of rigid motion of rotation having width f (Fig.2b) f   (20a) x x x y y y y x u u f y g f n n u u f x g f n n               (21b,c,d,e) x y x y x u c1 = y u c2 = f= f a) b) Figure 2 : Rigid motions: a) translation, b) rotation. Observations about the load vector as a consequence of the rigid motion 1) If the solid is subject to a rigid motion of translation (Fig.2a) having assigned values of the displacements c1 and 2 c , one has  U 0 and  G 0 but the total load vector has to be u g    L L L 0 with * , * , ( ) ( ) ut ut u tc u gt g tc                     A C a 0 L U 0 A a            (22) ( ) ur g gr gr                 A 0 L G A C 0  (23) and as a consequence the solution vector X has to be null. 2) If the solid is subject to a rigid motion of rotation (Fig.2a) having assigned rotation vector f  φ , in the generic node i one has i i    U r φ 0 i i i      G φ n φ s 0 (24a,b) with i r vector distance between the instantaneous center of rotation and the node i , i n e i s being respectively the unit normal and tangent vectors to the boundary elements on which the node i lies, but the total load vector has to be u g    L L L 0 with